Lemma 68.15.5. Let $S$ be a scheme. Let $f : X \to B$ and $B' \to B$ be morphisms of algebraic spaces over $S$. Assume

1. $B' \to B$ is a closed immersion,

2. $|B'| \to |B|$ is bijective,

3. $X \times _ B B' \to B'$ is a closed immersion, and

4. $X \to B$ is of finite type or $B' \to B$ is of finite presentation.

Then $f : X \to B$ is a closed immersion.

Proof. Assumptions (1) and (2) imply that $B_{red} = B'_{red}$. Set $X' = X \times _ B B'$. Then $X' \to X$ is closed immersion and $X'_{red} = X_{red}$. Let $U \to B$ be an étale morphism with $U$ affine. Then $X' \times _ B U \to X \times _ B U$ is a closed immersion of algebraic spaces inducing an isomorphism on underlying reduced spaces. Since $X' \times _ B U$ is a scheme (as $B' \to B$ and $X' \to B'$ are representable) so is $X \times _ B U$ by Lemma 68.15.3. Hence $X \to B$ is representable too. Thus we reduce to the case of schemes, see Morphisms, Lemma 29.44.7. $\square$

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